Question

Let $f:\left(-1,1\right)\to R$ be a differential function with $f\left(0\right)=-1$ and $f^{\prime }\left(0\right)=1$. Let $g\left(x\right)={\left[f\left(2f\left(x\right)+2\right)\right]}^2$

• A. $-4$
• B. $0$
• C. $-2$
• D. $4$

Answer 1

The correct option is A $-4$

Consider the given function.

$g\left(x\right)={\left[f(2f\left(x\right)+2)\right]}^2$

On differentiating both sides with respect to x, we get

$g^{\prime }\left(x\right)=2\left[f(2f\left(x\right)+2)\right]\times f^{\prime }(2f\left(x\right)+2)\times 2f^{\prime }\left(x\right)$

On putting x =0,

Therefore,

$g^{\prime }\left(x\right)=2\left[f(2f\left(0\right)+2)\right]\times f^{\prime }(2f\left(0\right)+2)\times 2f^{\prime }\left(0\right)$

Since, $f\left(0\right)=-1,f^{\prime }\left(0\right)=1$

Therefore,

$g^{\prime }\left(0\right)=2\left[f(2\times -1+2)\right]\times f^{\prime }(2\times -1+2)\times 2\times 1$

$g^{\prime }\left(0\right)=2\left[f\left(0\right)\right]\times f^{\prime }\left(0\right)\times 2$

$g^{\prime }\left(0\right)=2[-1]\times 1\times 2$

$g^{\prime }\left(0\right)=-2\times 1\times 2$

$g^{\prime }\left(0\right)=-4$

Hence, the value of $g^{\prime }\left(0\right)=-4$.